Solve it if you can. very easy 11th grade question

Part 1) {0,1,2,3,4}

Part 2) {February, April, June,September, November}

Part 3) {L, M, N, P, Q, R, S, T, V, W,X,Y, Z}

Step-by-step explanation:

Part 1) List the elements of the set

x is an integer for the interval (-1/2,9/2)

Remember that

-1/2=-0.5

9/2=4.5

so

we know that

The integers in the given interval are

(-0.5,4.5)

therefore

The elements of the set are

{0,1,2,3,4}

Part 2) List the elements of the set

x is an month of the year which does not contain 31 days

The months which not contains 31 days are

February, April, June,September, November

therefore

The elements of the set are

{February, April, June,September, November}

Part 3) List the elements of the set

x is a consonant of English alphabet which comes before k

we know that

The 21 consonant letters in the English alphabet are

B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W,X,Y, Z

so

The consonant of English alphabet which comes before k are

L, M, N, P, Q, R, S, T, V, W,X,Y, Z

therefore

The elements of the set are

{L, M, N, P, Q, R, S, T, V, W,X,Y, Z}

ok then what happened.

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.